Is the minimum and maximum of a set of admissible and consistent heuristics also consistent and admissible?
Yes, in both cases. Below I give two very simple proofs that directly follow from the definitions of admissible and consistent heuristics. However, in a nutshell, the idea of the proofs is that $h_{\max}(n)$ and $h_{\min}(n)$ are, by definition (of $h_{\max}$ and $h_{\min}$), equal to one of the given admissible (or consistent) heuristics, for all nodes $n$, so $h_{\max}(n)$ and $h_{\min}(n)$ are consequently admissible (or consistent).

Definitions

Consider the graph $G=(V, E, \mathcal{G})$ representing the search space, where $V$, $E$ and $\mathcal{G} \subseteq V$ are respectively the set of nodes, edges and goal nodes, and the function $w\colon E \times E \rightarrow \mathbb{R}$, which gives you the cost of each edge $e = (u, v) \in E$, where $u, v \in V$, that is, $w(e) = w(u, v) \in \mathbb{R}$ is the cost of the edge $e$.

A heuristic $h$ is admissible if $$h(n) \leq h^*(n), \forall n \in V,$$
where $h^*(n)$ is the optimal cost to reach a goal from node $n$ (that is, it is the optimal heuristic).

On the other hand, a heuristic $h$ is consistent if

\begin{align}
h(n) &amp;\leq w(n, s) + h(s), \forall n \in V \setminus \mathcal{G}, \text{ and} \\
h(n) &amp;= 0, \forall n \in \mathcal{G},
\end{align}
where $s$ is a successor of $n$.

Theorem 1

Given a set of admissible heuristics $H = \{ h_1, \dots, h_N \}$, then, for every $n \in V$, the heuristics $h_{\max}(n) = \max(h_1(n), \dots, h_N(n))$ and $h_{\min}(n) = \min(h_1(n), \dots, h_N(n))$ are also admissible.

Proof

Given that, $h_i(n) \leq h^*(n), \forall n \in V$ and $\forall i \in \{ 1, \dots N \}$, then $h_{\max}(n) = h_j(n) \leq h^*(n)$ (for some $j \in \{ 1, \dots N \}$)  and $h_{\min}(n) = h_k(n) \leq h^*(n)$ (for some $k \in \{ 1, \dots N \}$), so $h_{\max}$ and $h_{\min}$ are also admissible.

$\tag*{$\blacksquare$}$

Theorem 2

Given a set of consistent heuristics $H = \{ h_1, \dots, h_N \}$, then, for every $n \in V$, the heuristics $h_{\max}(n) = \max(h_1(n), \dots, h_N(n))$ and $h_{\min}(n) = \min(h_1(n), \dots, h_N(n))$ are also consistent.

Proof

Given that, for every $i \in \{ 1, \dots, N \}$, 

\begin{align}
\begin{cases}
h_i(n) \leq w(n, s) + h_i(s), &amp; \text{ if }  n \in V \setminus \mathcal{G}\\
h_i(n) = 0, &amp; \text{ if }  n \in \mathcal{G}
\end{cases}
\end{align}

then, if $n \in V \setminus \mathcal{G}$, $h_{\max}(n) = h_j(n) \leq w(n, s) + h_j(s)$, for some $j \in \{1, \dots, N \}$, and, similarly, $h_{\min}(n) = h_k(n) \leq w(n, s) + h_{k}(s)$, for some $k \in \{1, \dots, N \}$. Similarly, if $n \in \mathcal{G}$, $h_{\max}(n) = h_{\min}(n) = h_i(n) = 0, \forall i \in \{1, \dots, N \} $. Thus, $h_{\max}$ and $h_{\min}$ are also consistent.

$\tag*{$\blacksquare$}$

Notes

$h_{\max}$ and $h_{\min}$ have been defined in the only possible reasonable way, because, given two (admissible or consistent) heuristics, one may not always dominate the other (or vice-versa), even if both are admissible and consistent.